Chapter 5: Problem 68
Let \(X\) be a binomial random variable with \(p=0.25\) and \(n=3\). Find the probability distribution of the random variable \(Y=X^{2}\).
Short Answer
Expert verified
\(P(Y=0) = 0.421875\), \(P(Y=1) = 0.421875\), \(P(Y=4) = 0.140625\), \(P(Y=9) = 0.015625\).
Step by step solution
01
Identify the Probability Distribution
The random variable \(X\) is binomially distributed with \(n=3\) and \(p=0.25\). Thus, \(X\) can take values \(0, 1, 2,\) and \(3\).
02
Calculate Binomial Probabilities
We calculate \(P(X=k)\) using the binomial formula: \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\).\- \(P(X=0) = \binom{3}{0} (0.25)^0 (0.75)^3 = 0.421875\)\- \(P(X=1) = \binom{3}{1} (0.25)^1 (0.75)^2 = 0.421875\)\- \(P(X=2) = \binom{3}{2} (0.25)^2 (0.75)^1 = 0.140625\)\- \(P(X=3) = \binom{3}{3} (0.25)^3 (0.75)^0 = 0.015625\)
03
Find the Values of Y
Since \(Y = X^2\), the possible values \(Y\) can take are determined by squaring the values of \(X\): \(0^2 = 0\), \(1^2 = 1\), \(2^2 = 4\), \(3^2 = 9\). Thus, \(Y\) can take the values \(0, 1, 4,\) and \(9\).
04
Determine the Probability Mass Function of Y
To find \(P(Y=y)\) for each \(y\), we sum the probabilities of \(X\) values that give \(Y=y\).\- \(P(Y=0) = P(X=0) = 0.421875\)\- \(P(Y=1) = P(X=1) = 0.421875\)\- \(P(Y=4) = P(X=2) = 0.140625\)\- \(P(Y=9) = P(X=3) = 0.015625\)
05
Construct the Probability Distribution of Y
The probability distribution of \(Y\) is as follows: \(P(Y=0) = 0.421875\), \(P(Y=1) = 0.421875\), \(P(Y=4) = 0.140625\), and \(P(Y=9) = 0.015625\). This gives the complete distribution of \(Y\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability Mass Function
The Probability Mass Function, often abbreviated as PMF, is a crucial concept when dealing with discrete random variables. In probability theory, a PMF assigns probabilities to all possible outcomes of a discrete random variable. This function, mathematically expressed as \( P(X=k) \), describes the likelihood of a random variable \( X \) taking on a particular value \( k \).
In essence, the PMF is a vital tool for understanding the distribution of a discrete random variable's probabilities.
- For instance, in the given problem, \( X \) is a binomial random variable that can assume the values \( 0, 1, 2, \) and \( 3 \).
- Each of these potential outcomes is associated with a calculated probability using the Binomial PMF formula.
In essence, the PMF is a vital tool for understanding the distribution of a discrete random variable's probabilities.
Random Variable
A Random Variable is a variable whose possible values are numerical outcomes of a random process. It can be thought of as a way to quantify outcomes of random phenomena numerically.
In our scenario, we have the random variable \( X \) which is governed by the binomial distribution parameters \( n=3 \) and \( p=0.25 \). The possible values that \( X \) can take are \( 0, 1, 2, \) and \( 3 \), each associated with a particular probability.
In our scenario, we have the random variable \( X \) which is governed by the binomial distribution parameters \( n=3 \) and \( p=0.25 \). The possible values that \( X \) can take are \( 0, 1, 2, \) and \( 3 \), each associated with a particular probability.
- These outcomes occur due to the random nature of the binomial experiment where each trial is independent, and there is a fixed probability of success (\( p=0.25 \)).
- The square of \( X \), denoted \( Y=X^2 \), represents another random variable derived from \( X \) and has its own possible outcomes \( 0, 1, 4, \) and \( 9 \).
Probability Distribution
Probability Distribution refers to the mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. For a discrete random variable, this entails listing the potential values the random variable can assume, alongside their respective probabilities.
In our example, the probability distribution of the random variable \( Y \) can be summarized as follows:
Each probability indicates how likely a particular squared outcome is to occur in the process described by the random experiment.
In our example, the probability distribution of the random variable \( Y \) can be summarized as follows:
- \( P(Y=0) = 0.421875 \)
- \( P(Y=1) = 0.421875 \)
- \( P(Y=4) = 0.140625 \)
- \( P(Y=9) = 0.015625 \)
Each probability indicates how likely a particular squared outcome is to occur in the process described by the random experiment.
Binomial Random Variable
A Binomial Random Variable is a specific type of random variable that arises from a series of repeated, independent trials of a binary outcome process, such as flipping a coin. The binomial distribution is determined by two parameters: \( n \), the number of trials, and \( p \), the probability of success in each trial.
For example, in our exercise, \( X \) is a binomial random variable with parameters \( n=3 \) and \( p=0.25 \). This means we conduct three trials, each with a 25% probability of success.
For example, in our exercise, \( X \) is a binomial random variable with parameters \( n=3 \) and \( p=0.25 \). This means we conduct three trials, each with a 25% probability of success.
- The possible outcomes for \( X \) can range from 0 successes (none) to 3 successes (all trials are successful).
- These outcomes follow a pattern dictated by the binomial probabilities derived from the formula: \( P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\).